Hi everyone!
I am currently studying the basic theory of measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost...

Question: Suppose $T$ is an invertible
measure-preserving map of standard probability measure
space $(X,\mu)$. Suppose that $TA=A$
for all measurable subsets
$A\subset{}X$, where the equality is
up to sets of measure $0$. Prove that
the set of those $x$ where $Tx\neq{}x$
has measure $0$.

You will need some countability-type hypotheses on the measure space. For example, it is a standard measure space. If not, build a counterexample in product space $[0,1]^{[0,1]}$.
–
Gerald EdgarApr 25 '11 at 13:20

2 Answers
2

If $X$ is a standard probability space then we may assume it to be the disjoint union of an interval with Lebesgue measure and a countable set of atoms. If $p$ is an atom, then by assumption, $T(p) = p$, since $p$ has positive measure. So none of the atoms can be "bad" points. So we may assume that there are no atoms, so that $X=[0,1]$ and the measure is Lebesgue measure. Now consider all subintervals $I$ of $[0,1]$ with rational endpoints and gather all points in $TI$ which are outside $I$. We get a countable union of measure zero sets, hence a measure zero set. Denote it by $Z$. Now let $x \notin Z$. This implies that $Tx$ belongs to arbitrarily small intervals around $x$ and is thus equal to $x$. The desired result follows.

not an answer to the revised question
But Das Curious asks for the counterexample (to the original question) when the countability assumption is omitted. Here it is.

Let $X = [0,1)^{[0,1]}$, that is: the set of all functions $x \colon [0,1] \to [0,1)$. Let $\lambda$ be Lebesgue measure on $[0,1)$. Let $\Lambda$ be the product measure on $X$ with the product $\sigma$-algebra $\mathcal F$. In particular: if $A \in \mathcal F$, then there is a countable $J \subseteq [0,1]$ such that $A$ "depends only on" $J$ in the sense that: for $x,y\in X$ with $x(t)=y(t)$ for all $t \in J$, then either $x,y$ both belong to $A$ or both belong to $X \setminus A$.

For $s \in [0,1)$ let $Q_s = \{x \in X\;|\; x(1)=s\}$. Thus $(Q_s)_{s \in [0,1)}$ is a family of subsets of $X$; the family has cardinal of the continuum, so it is indexed by $[0,1)$; each $Q_s$ is a measurable set with $\Lambda(Q_s)=0$; the sets $Q_s$ are pairwise disjoint; but the union $\bigcup_{s} Q_s = X$ is the whole space.

Now we can define $T \colon X \to X$. Let
$$
T(x)(t) = \theta(x(t))\qquad\text{if } x \in Q_{t}
\\
T(x)(t) = x(t)\qquad\text{if }x \notin Q_t
$$
Now each $x$ belongs to exactly one $Q_t$, so $x$ and $T(x)$ agree, except on a single coordinate, where they disagree. Note $T(x) \ne x$ for all $x$. Because $1 \notin [0,1)$, the map $T$ maps each $Q_s$ into itself. Thus $T$ is a bijection of $X$ onto itself: the inverse map has the same definition using $\theta^{-1}$ in place of $\theta$.

We claim that the map $T$ is measurable. Let $A \in \mathcal F$. We claim $T^{-1}(A) \in \mathcal F$. There is a countable set $J \subseteq [0,1)$ such that $A$ depends only on $J$. For $s \in J$ let $A_s := A \cap Q_s$. Let $B := A \setminus \bigcup_{s \in J} Q_s$. Now $A$ is the countable union
$$
A = B \cup \bigcup_{s \in J} A_s
$$
where $B$ and all $A_s$ belong to $\mathcal F$.
The inverse image $T^{-1}(A)$ is the countable union
$$
T^{-1}(A) = T^{-1}(B) \cup \bigcup_{s \in J} T^{-1}(A_s) .
$$
Now by the definition of $J$, we have $T^{-1}(B)=B$. Each $\theta_s$ is a measurable function, and
$T^{-1}(A_s) = \theta_s^{-1}(A_s)$, so each $T^{-1}(A_s) \in \cal F$. Therefore $T^{-1}(A)$ is written as a countable union of sets from $\mathcal F$.

We claim $A = T(A)$ up to null sets for any $A \in \mathcal F$.
Let $J$ be as before.
If $x$ is not in the null set $\bigcup_{s \in J} Q_s$, then either $x,T(x)$ are both in $A$ or both in $X \setminus A$. So $A$ and $T(A)$ agree except for adding and subtracting parts of that null set.